according to the nyquist theorem, the sampling rate must be
DESCRIPTION
Note. According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency contained in the signal. Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals. Example 4.6. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/1.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
According to the Nyquist theorem, the sampling rate must be
at least 2 times the highest frequency contained in the signal.
Note
![Page 2: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/2.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.23 Nyquist sampling rate for low-pass and bandpass signals
![Page 3: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/3.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
For an intuitive example of the Nyquist theorem, let us sample a simple sine wave at three sampling rates: fs = 4f (2 times the Nyquist rate), fs = 2f (Nyquist rate), and fs = f (one-half the Nyquist rate). Figure 4.24 shows the sampling and the subsequent recovery of the signal.
It can be seen that sampling at the Nyquist rate can create a good approximation of the original sine wave (part a). Oversampling in part b can also create the same approximation, but it is redundant and unnecessary. Sampling below the Nyquist rate (part c) does not produce a signal that looks like the original sine wave.
Example 4.6
![Page 4: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/4.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.24 Recovery of a sampled sine wave for different sampling rates
![Page 5: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/5.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Telephone companies digitize voice by assuming a maximum frequency of 4000 Hz. The sampling rate therefore is 8000 samples per second.
Example 4.9
![Page 6: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/6.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
A complex low-pass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?
SolutionThe bandwidth of a low-pass signal is between 0 and f, where f is the maximum frequency in the signal. Therefore, we can sample this signal at 2 times the highest frequency (200 kHz). The sampling rate is therefore 400,000 samples per second.
Example 4.10
![Page 7: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/7.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
A complex bandpass signal has a bandwidth of 200 kHz. What is the minimum sampling rate for this signal?
SolutionWe cannot find the minimum sampling rate in this case because we do not know where the bandwidth starts or ends. We do not know the maximum frequency in the signal.
Example 4.11
![Page 8: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/8.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.26 Quantization and encoding of a sampled signal
![Page 9: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/9.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
What is the SNRdB in the example of Figure 4.26?
SolutionWe can use the formula to find the quantization. We have eight levels and 3 bits per sample, so
SNRdB = 6.02(3) + 1.76 = 19.82 dB
Increasing the number of levels increases the SNR.
Example 4.12
![Page 10: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/10.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
A telephone subscriber line must have an SNRdB above 40. What is the minimum number of bits per sample?
SolutionWe can calculate the number of bits as
Example 4.13
Telephone companies usually assign 7 or 8 bits per sample.
![Page 11: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/11.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample?
SolutionThe human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows:
Example 4.14
![Page 12: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/12.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.27 Components of a PCM decoder
![Page 13: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/13.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
We have a low-pass analog signal of 4 kHz. If we send the analog signal, we need a channel with a minimum bandwidth of 4 kHz. If we digitize the signal and send 8 bits per sample, we need a channel with a minimum bandwidth of 8 × 4 kHz = 32 kHz.
Example 4.15
![Page 14: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/14.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
4-3 TRANSMISSION MODES4-3 TRANSMISSION MODES
The transmission of binary data across a link can be The transmission of binary data across a link can be accomplished in either parallel or serial mode. In accomplished in either parallel or serial mode. In parallel mode, multiple bits are sent with each clock parallel mode, multiple bits are sent with each clock tick. In serial mode, 1 bit is sent with each clock tick. tick. In serial mode, 1 bit is sent with each clock tick. While there is only one way to send parallel data, there While there is only one way to send parallel data, there are three subclasses of serial transmission: are three subclasses of serial transmission: asynchronous, synchronous, and isochronous.asynchronous, synchronous, and isochronous.
Parallel TransmissionSerial Transmission
Topics discussed in this section:Topics discussed in this section:
![Page 15: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/15.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.31 Data transmission and modes
![Page 16: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/16.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.32 Parallel transmission
![Page 17: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/17.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.33 Serial transmission
![Page 18: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/18.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
In asynchronous transmission, we send 1 start bit (0) at the beginning and 1 or more stop bits (1s) at the end of each
byte. There may be a gap between each byte.
Note
![Page 19: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/19.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Asynchronous here means “asynchronous at the byte level,”
but the bits are still synchronized; their durations are the same.
Note
![Page 20: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/20.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.34 Asynchronous transmission
![Page 21: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/21.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
In synchronous transmission, we send bits one after another without start or
stop bits or gaps. It is the responsibility of the receiver to group the bits.
Note
![Page 22: According to the Nyquist theorem, the sampling rate must be](https://reader036.vdocuments.net/reader036/viewer/2022081506/56814ffc550346895dbdc35c/html5/thumbnails/22.jpg)
Dept. of EEE, RUET A. B. M. Nasiruzzaman
Figure 4.35 Synchronous transmission